Given that 𝑓(𝑥) = 3𝑥 2 − 4𝑥 + 7, use the definition of the derivative to find 𝑓 ′ (𝑥).
Given that 𝑓(𝑥)=3𝑥2−4𝑥+7𝑓(𝑥) = 3𝑥^2 − 4𝑥 + 7f(x)=3x2−4x+7 , let us find 𝑓′(𝑥)𝑓'(𝑥)f′(x):
f′(x)=limΔx→0f(x+Δx)−f(x)Δx=limΔx→03(x+Δx)2−4(x+Δx)+7−(3x3−4x+7)Δx=limΔx→03x3+6xΔx+3(Δx)2−4x−4Δx+7−3x2+4x−7Δx=limΔx→06xΔx+3(Δx)2−4ΔxΔx=limΔx→0(6x+3Δx−4)=6x−4.f'(x)=\lim\limits_{\Delta x\to 0}\frac{f(x+\Delta x)-f(x)}{\Delta x}= \lim\limits_{\Delta x\to 0}\frac{3(x+\Delta x)^2-4(x+\Delta x)+7-(3x^3-4x+7)}{\Delta x}= \lim\limits_{\Delta x\to 0}\frac{3x^3+6x\Delta x+3(\Delta x)^2-4x-4\Delta x+7-3x^2+4x-7}{\Delta x}= \lim\limits_{\Delta x\to 0}\frac{6x\Delta x+3(\Delta x)^2-4\Delta x}{\Delta x}= \lim\limits_{\Delta x\to 0}(6x+3\Delta x-4)=6x-4.f′(x)=Δx→0limΔxf(x+Δx)−f(x)=Δx→0limΔx3(x+Δx)2−4(x+Δx)+7−(3x3−4x+7)=Δx→0limΔx3x3+6xΔx+3(Δx)2−4x−4Δx+7−3x2+4x−7=Δx→0limΔx6xΔx+3(Δx)2−4Δx=Δx→0lim(6x+3Δx−4)=6x−4.
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